用定义法求导数问题y=根号下x在x=1处的导数

定义法y = √x y'|(x = x0) = lim(△x→0) △y/△x = lim(x→x0) (y-y0)/(x -x0)对于 x0 = 1y'|(x=1) = lim(x→1) (√x - √1)/(x - 1)= lim(x→1) (√x -1)/[(√x + 1)(√x -1)]= lim(x→1) 1/(√x + 1)= 1/(√1 + 1)= 1/2公式法y = √x = x^(1/2)y' = (1/2) * x^(1/2 -1) = (1/2)*x^(-1/2) = 1/(2√x)在 x = 1 处y'(x=1) = 1/(2√1) = 1/2